Comments on: Question About The Infinite Sum 1/k Equal To Infinite Product Of 1/(1-1/p)… If You Are Familiar With This?

By: Curt Monash

Curt Monash — Thu, 09 Jul 2009 06:32:26 +0000

Look at the product for 2, 3, 5, … up to the nth prime.
Compare that to the sum over, 1, 2, 3, … up to, not the nth prime, but rather the product of the first n primes. See where you are.
If that doesn’t work well, try some similar idea.

By: zpconn

zpconn — Thu, 09 Jul 2009 03:50:32 +0000

If the sum and the product diverge, what does it mean for them to become equal at “the infinitieth k” and at “the infinitieth prime”? Further, what is the “infinite point” of a sequence, convergent or divergent?
Perhaps you’re coming to this question because you’ve read that the fact that the Riemann zeta function has a simple pole at z = 1 is a reflection of the fact that there are infinitely many primes.
The argument works out like this. Suppose temporarily that there are finitely many primes. Working through the algebra, we find that this particular product–which is nothing more than a number since we’re supposing a finite number of primes–must be equal to the sum of the harmonic series. However, the harmonic series diverges, so it can’t be equal to a number. This is a contradiction; therefore, we conclude that there are infinitely many primes.
More rigorously, one can show that the harmonic series converges if and only if the product of primes converges. Hence, the divergence of the harmonic series implies the divergence of the product of primes. As you probably know, this implies that there are infinitely many primes.
What Sean H said is, despite your having scolded him, accurate. The equality in this form has no meaning other than that both limits diverge. In fact, this equality is often directly interpreted to mean “the prime product diverges because the harmonic series diverges.” It’s written like for several reasons:
1) It’s suggestive and intuitive; many find the fact that the harmonic series diverges implies there are infinitely many primes a truly wonderful argument.
2) It’s historical; Euler wrote it this way.
3) It’s a special case of a more general result in which the equality is far more meaningful. Specifically, for real s > 1 it can be shown that
Zeta(s) = ∏_{prime p} 1 / [1 − 1/(p^s)]
where Zeta(s) is the Riemann zeta function. In this case, the left and right sides are considered as functions of s (for s > 1), and the equality states that not only do both converge for such s, they converge to the same value. Taking this result and plugging in s = 1 yields your formula. Of course, the zeta function has a pole (i.e., the series diverges) at s = 1, so the equality loses much of its meaning, but people still tend to do this because, as I mentioned above, it can be shown that if the harmonic series diverges then the prime product must also diverge.

By: Sean H

Sean H — Thu, 09 Jul 2009 02:31:23 +0000

The equality merely asserts that they both diverge, or that both limits are positive infinity. No more, no less. Don’t try to make it more than it is.
Many problems come up when you try to do arithmetic with infinity. There are some ways to work it out, but I don’t think this particular equality has much to do with that. As I said above, all it asserts is that both sides diverge to infinity. It is similar to the statement:
lim_{x->infty} x = lim_{x->infty} x^2.
Edit: Unless you have some proof that I’m not familiar with, what is actually shown is that:
Sum_{k=1}^infty 1/(k^s) = Product_{p prime} 1/(1-1/p^s)
for Re(s) > 1. Then, taking s -> 1 along the real axis you can show that Product_{p prime} 1/(1-1/p) diverges, which is the significance of the formula about which you are asking. As cpconn pointed out, this implies that there must be infinitely many prime numbers.